\documentclass{article}
\newtheorem{axiom}{Axiom}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\newtheorem{proposition}{Proposition}
\newtheorem{definition}{Definition}
\newtheorem{example}{Example}
\newtheorem{conjecture}{Conjecture}
\newcommand\unicdcdgdad{\not=}
\newcommand\unicdcdcdbg{\cup}
\newcommand\unicdcdbdgd{\setminus}
\newcommand\unicdcdcdjd{\cap}
\newcommand\unicdcdadad{\forall}
\newcommand\unicdbdjdcd{\Rightarrow}
\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
\newcommand\unicdcdcdid{\lor}
\newcommand\unicdcdcdhd{\land}
\newcommand\uniddcgcg{\lambda}
\newcommand\unicdcdaddd{\exists}
\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{l26\_dtconstr (TMP9EFj6sQsCSVFZFCdmvYVobmVUnKbzuX7)}
\maketitle
Let $ k2\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k5\_dtconstr : {\iota}$ be given.
Let $ k6\_domain\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ u1\_struct\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ c3\_\_dtconstr : {\iota}$ be given.
Let $ r1\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}r1\_tarski~(k2\_lang1~k5\_dtconstr)~(k6\_domain\_1~(u1\_struct\_0~k5\_dtconstr)~\\
c3\_\_dtconstr)\end{array}
\label{hyp:mgcdfdpfegehdgpgogdhehch}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}r1\_tarski~(k6\_domain\_1~(u1\_struct\_0~k5\_dtconstr)~c3\_\_dtconstr)~\\
(k2\_lang1~k5\_dtconstr)\end{array}
\label{hyp:mgcdedpfegehdgpgogdhehch}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{({{X0}={X1}})}{\unicdbdjded}{({{({{{r1\_tarski}~{X0}}~{X1}})}{\unicdcdcdhd}{({{{r1\_tarski}~{X1}}~{X0}})}})}}}}\label{hyp:egbdadpfihcgpgpgmgfgpfad}
\end{equation}
\begin{theorem}\label{thm:mgcdgdpfegehdgpgogdhehch}
$$\begin{array}{c}k2\_lang1~k5\_dtconstr=k6\_domain\_1~(u1\_struct\_0~k5\_dtconstr)~\\
c3\_\_dtconstr\end{array}$$
\end{theorem}
\end{document}
